Question

A tangent to the curve $y=x^2+3x$ passes through a point $(0,-9)$ if it drawn at the point-

• A. $(-3,0)$
• B. $(1,4)$
• C. $(0,0)$
• D. $(-4,4)$

Answer 1

Answer: (A)

$(-3,0)$

Consider the equation of the curve.

$y=x^2+3x$

Slope of the tangent to this curve at any point $(h,k)$ is,

$\frac{dy}{dx}=2x+3$

${\frac{dy}{dx}]}_{x=h,y=k}=2h+3$

Therefore, equation of the tangent will be,

$y-y_1=(2x+3)(x-x_1)$

Since, the tangent passes through $(0,-9)$, so

$y+9=(2x+3)\left(x\right)$

$y+9=2hx+3x$

$y=2hx+3x-9$

Now, since the point $(h,k)$ lies both on the curve and the tangent, we have

$h^2+3h=2h^2+3h-9$

$h^2=9$

$h=\pm 3$

Therefore, the points are,

$y_1={\left(3\right)}^2+3\left(3\right)=18$

$y_2={(-3)}^2+3(-3)=0$

Therefore, the required points are,

$\Rightarrow (3,18)$ and $(-3,0)$

Hence, this is the required answer.